The period of revolution of planet A around t | vedclass

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(C) According to Kepler's third law of planetary motion,the square of the period of revolution $T$ is proportional to the cube of the semi-major axis

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) According to Kepler's third law of planetary motion,the square of the period of revolution $T$ is proportional to the cube of the semi-major axis $r$ of the orbit: $T^2 \propto r^3$.Given that the period of revolution of planet $A$ is $8$ times that of planet $B$,we have $T_A = 8 T_B$,or $\frac{T_A}{T_B} = 8$.Using the relation $\frac{T_A}{T_B} = \left( \frac{r_A}{r_B} \right)^{3/2}$,we substitute the given values:$8 = \left( \frac{r_A}{r_B} \right)^{3/2}$.To solve for $\frac{r_A}{r_B}$,we raise both sides to the power of $2/3$:$\left( 8 \right)^{2/3} = \frac{r_A}{r_B}$.Since $8 = 2^3$,we have $(2^3)^{2/3} = 2^2 = 4$.Therefore,$\frac{r_A}{r_B} = 4$,which means the distance of planet $A$ from the sun is $4$ times greater than that of planet $B$.

The period of revolution of planet $A$ around the sun is $8$ times that of planet $B$. The distance of planet $A$ from the sun is how many times greater than that of planet $B$ from the sun?

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